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28x^{2}-8x-48=0
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 28\left(-48\right)}}{2\times 28}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 28 for a, -8 for b, and -48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-8\right)±\sqrt{64-4\times 28\left(-48\right)}}{2\times 28}
Square -8.
x=\frac{-\left(-8\right)±\sqrt{64-112\left(-48\right)}}{2\times 28}
Multiply -4 times 28.
x=\frac{-\left(-8\right)±\sqrt{64+5376}}{2\times 28}
Multiply -112 times -48.
x=\frac{-\left(-8\right)±\sqrt{5440}}{2\times 28}
Add 64 to 5376.
x=\frac{-\left(-8\right)±8\sqrt{85}}{2\times 28}
Take the square root of 5440.
x=\frac{8±8\sqrt{85}}{2\times 28}
The opposite of -8 is 8.
x=\frac{8±8\sqrt{85}}{56}
Multiply 2 times 28.
x=\frac{8\sqrt{85}+8}{56}
Now solve the equation x=\frac{8±8\sqrt{85}}{56} when ± is plus. Add 8 to 8\sqrt{85}.
x=\frac{\sqrt{85}+1}{7}
Divide 8+8\sqrt{85} by 56.
x=\frac{8-8\sqrt{85}}{56}
Now solve the equation x=\frac{8±8\sqrt{85}}{56} when ± is minus. Subtract 8\sqrt{85} from 8.
x=\frac{1-\sqrt{85}}{7}
Divide 8-8\sqrt{85} by 56.
x=\frac{\sqrt{85}+1}{7} x=\frac{1-\sqrt{85}}{7}
The equation is now solved.
28x^{2}-8x-48=0
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.
28x^{2}-8x-48-\left(-48\right)=-\left(-48\right)
Add 48 to both sides of the equation.
28x^{2}-8x=-\left(-48\right)
Subtracting -48 from itself leaves 0.
28x^{2}-8x=48
Subtract -48 from 0.
\frac{28x^{2}-8x}{28}=\frac{48}{28}
Divide both sides by 28.
x^{2}+\left(-\frac{8}{28}\right)x=\frac{48}{28}
Dividing by 28 undoes the multiplication by 28.
x^{2}-\frac{2}{7}x=\frac{48}{28}
Reduce the fraction \frac{-8}{28} to lowest terms by extracting and canceling out 4.
x^{2}-\frac{2}{7}x=\frac{12}{7}
Reduce the fraction \frac{48}{28} to lowest terms by extracting and canceling out 4.
x^{2}-\frac{2}{7}x+\left(-\frac{1}{7}\right)^{2}=\frac{12}{7}+\left(-\frac{1}{7}\right)^{2}
Divide -\frac{2}{7}, the coefficient of the x term, by 2 to get -\frac{1}{7}. Then add the square of -\frac{1}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{2}{7}x+\frac{1}{49}=\frac{12}{7}+\frac{1}{49}
Square -\frac{1}{7} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{2}{7}x+\frac{1}{49}=\frac{85}{49}
Add \frac{12}{7} to \frac{1}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{1}{7}\right)^{2}=\frac{85}{49}
Factor x^{2}-\frac{2}{7}x+\frac{1}{49}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x-\frac{1}{7}\right)^{2}}=\sqrt{\frac{85}{49}}
Take the square root of both sides of the equation.
x-\frac{1}{7}=\frac{\sqrt{85}}{7} x-\frac{1}{7}=-\frac{\sqrt{85}}{7}
Simplify.
x=\frac{\sqrt{85}+1}{7} x=\frac{1-\sqrt{85}}{7}
Add \frac{1}{7} to both sides of the equation.
x ^ 2 -\frac{2}{7}x -\frac{12}{7} = 0
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 28
r + s = \frac{2}{7} rs = -\frac{12}{7}
Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C
r = \frac{1}{7} - u s = \frac{1}{7} + u
Two numbers r and s sum up to \frac{2}{7} exactly when the average of the two numbers is \frac{1}{2}*\frac{2}{7} = \frac{1}{7}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>
(\frac{1}{7} - u) (\frac{1}{7} + u) = -\frac{12}{7}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{12}{7}
\frac{1}{49} - u^2 = -\frac{12}{7}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{12}{7}-\frac{1}{49} = -\frac{85}{49}
Simplify the expression by subtracting \frac{1}{49} on both sides
u^2 = \frac{85}{49} u = \pm\sqrt{\frac{85}{49}} = \pm \frac{\sqrt{85}}{7}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{1}{7} - \frac{\sqrt{85}}{7} = -1.174 s = \frac{1}{7} + \frac{\sqrt{85}}{7} = 1.460
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.